(a) A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hours duration with intensity of 70 mm/h in the first hour and 40 mm/h in the second hour occurs over the catchment. Calculate the peak direct surface runoff value i

Question

(a) A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hours duration with intensity of 70 mm/h in the first hour and 40 mm/h in the second hour occurs over the catchment. Calculate the peak direct surface runoff value in m³/s. (10 marks)

(b) A temporary coffer dam is to be built to protect the 5 year construction activity for a major river valley project. If the coffer dam is designed to withstand the 25-year flood, what is the risk that the structure will be overtopped in the third year? (10 marks)

(c) A weir across an alluvial river has a horizontal floor length of 50 m and retains 5 m of water under full flow condition. The downstream sheet pile is driven to a depth of 5 m below the impervious floor of negligible thickness. Determine the exit gradient. (10 marks)

(d) What is an indicator organism ? Discuss the required characteristics for an ideal indicator organism. Name any two indicator organisms. (10 marks)

(e) Circular sewer of minimum diameter of 150 mm is recommended for the house connections. Calculate the discharge carried by this sewer, when flowing full, laid at a slope of 0·004. Take Manning's n as 0·013.

Why this sewer size has been recommended as the minimum size, when the waste water discharge from individual household will be substantially less ? (10 marks)

UPSC Answer Check · Accepted Answer

(a) Method: unit-hydrograph convolution. Since no losses are given, the given rainfall intensities are taken as effective rainfall excess. The 1-hour unit hydrograph ordinates are:
U(0)=0 m³/s per cm
U(1)=3 m³/s per cm
U(2)=2 m³/s per cm
U(3)=1 m³/s per cm
U(4)=0 m³/s per cm

Storm rainfall:
First hour: 70 mm = 7 cm
Second hour: 40 mm = 4 cm

Convolution ordinates:
t=1 h: 7×3 = 21 m³/s
t=2 h: 7×2 + 4×3 = 14+12 = 26 m³/s
t=3 h: 7×1 + 4×2 = 7+8 = 15 m³/s
t=4 h: 7×0 + 4×1 = 4 m³/s
t=5 h: 0

Continuous check: for 1≤t≤2 h, Q(t)=7(4−t)+4[3(t−1)]=16+5t, which increases up to t=2 h. For 2≤t≤3 h, Q(t)=7(4−t)+4(5−t)=48−11t, which decreases. Hence the maximum occurs at t=2 h.

Peak direct surface runoff = 26 m³/s at t = 2 h after storm start.

(b) Method: annual exceedance probability for a return period T.

For T=25 years,

p = 1/T = 1/25 = 0·04

Assuming independent annual flood events, the probability that the coffer dam is overtopped during the third year alone is the same as the annual exceedance probability.

Risk of overtopping in the third year = 0·04 = 4%.

If the question is interpreted as cumulative risk by the third year, then:
Risk = 1 − (1−p)³ = 1 − (0·96)³ = 1 − 0·884736 = 0·115264 = 11·5264%.

If interpreted as cumulative risk over the full 5-year construction period:
Risk = 1 − (0·96)⁵ = 0·184627 = 18·4627%.

Since the wording asks “in the third year”, the answer is 4%.

(c) Method: Khosla’s exit-gradient formula for a downstream sheet pile.

For downstream sheet pile depth d, impervious floor length b, and head H,

G_E = (H/d) × 1/(π√λ)

λ = (1 + √(1+α²))/2

α = b/d

Given:
H = 5 m, b = 50 m, d = 5 m.

α = 50/5 = 10

λ = (1 + √(1+10²))/2 = (1+√101)/2

√101 = 10·0498756

λ = (1+10·0498756)/2 = 5·5249378

√λ = √5·5249378 = 2·350518

G_E = (5/5) × 1/(π×2·350518)

G_E = 1/(7·38437)

G_E = 0·13542

Exit gradient = 0·1354 (dimensionless).

Condition of validity: floor thickness is negligible, downstream sheet pile is at the end of the impervious floor, and the formula applies to the Khosla exit-gradient condition in alluvial soil.

(d) An indicator organism is a microorganism whose presence in water indicates probable faecal contamination and therefore the possible presence of pathogenic bacteria, viruses, or protozoa. It is used because direct testing for every pathogen is difficult, costly, and slow.

Required characteristics of an ideal indicator organism:
It should be present in faecal material in large numbers.
It should be absent, or nearly absent, in uncontaminated water.
It should not multiply outside the host, so its count reflects the degree of pollution.
It should survive in water at least as long as the pathogens of concern.
It should be more resistant to disinfection than pathogens, so its absence gives a margin of safety.
It should be easily, rapidly, and economically detected by simple laboratory methods.
It should be non-pathogenic or of very low pathogenicity to humans.
It should be specific to faecal pollution and not originate from non-faecal sources.
It should have stable and uniform characteristics in different water environments.

Two indicator organisms are:
Escherichia coli
Faecal streptococci, e.g. Enterococcus faecalis

(e)(i) Method: Manning’s formula for a circular sewer flowing full.

Given:
D = 150 mm = 0·150 m
S = 0·004
n = 0·013

For a circular pipe flowing full,

A = πD²/4 = π(0·150)²/4 = 0·0176715 m²

P = πD = π×0·150 = 0·471239 m

R = A/P = D/4 = 0·150/4 = 0·0375 m

R^(2/3) = (0·0375)^(2/3) = 0·112035

S^(1/2) = √0·004 = 0·0632456

Manning’s equation:

Q = (1/n) A R^(2/3) S^(1/2)

Q = (1/0·013)(0·0176715)(0·112035)(0·0632456)

Q = 0·009632 m³/s

Discharge when flowing full = 0·009632 m³/s = 9·632 L/s ≈ 9·63 L/s.

Condition: sewer is flowing full, flow is steady and uniform, and Manning’s n = 0·013.

(e)(ii) A 150 mm diameter is recommended as the minimum size for house connections mainly for reasons other than the average household discharge:
It prevents frequent choking due to sanitary solids, paper, rags, and other domestic wastes.
Smaller diameters, such as 100 mm, have a high risk of blockage at bends, junctions, and low-flow periods.
It provides reserve hydraulic capacity for peak household flows, infiltration, and future additions.
It allows easy cleaning, rodding, and inspection of the sewer line.
It helps maintain reasonable self-cleansing conditions during peak flow and reduces long stagnation of sewage.
It permits ventilation and safe movement of sewage gases.
It is a standard minimum requirement in sanitary codes and practice, ensuring durability, maintainability, and reliable operation.

Thus, the minimum size is fixed by clogging avoidance and maintenance requirements, not by the small average discharge from an individual household.

(a) A one-hour unit hydrograph of a catchment is shown in the figure. A storm of two hours duration with intensity of 70 mm/h in…

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